Force, Mass and Momentum - Rocoscience



Chapter 9: Force, Mass and Momentum

Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier

A Force is anything which can cause an object to accelerate.

The unit of force is the Newton (N)*.

A force of 1 N gives a mass of 1 kg an acceleration of 1 m s-1.

The Mass of an object is a measure of how difficult it is to accelerate that object.

Or

The Mass of an object is a measure of its Inertia.

(The inertia of an object in turn is a measure of how difficult it is to accelerate it.)

The unit of mass is the kilogram (kg).

Relationship between Force, Mass and Acceleration

Force = mass × acceleration

Always remember: “AN UNBALANCED FORCE PRODUCES AN ACCELERATION”

Weight

The weight of an object is a measure of the force of the Earth’s gravity acting on it.

Weight = mass × acceleration due to gravity

Because weight is a force, it follows that the unit of weight is also the Newton.

Friction

Friction is a force which opposes the relative motion between two objects.

Examples of Friction: Brakes, Walking, Air Resistance

Momentum

Momentum = Mass × Velocity

The symbol of momentum is (; pronounced “row”)

The unit of Momentum is the kilogram metre per second (kg m s-1)

Newton’s Laws of Motion

1. Newton’s First Law of Motion states that every object will remain in a state of rest or travelling with a constant velocity unless an external force acts on it.

2. Newton’s Second Law of Motion states that the rate of change of an object’s momentum is directly proportional to the force which caused it, and takes place in the direction of the force.

3. Newton’s Third Law of Motion* states that when body A exerts a force on body B, B exerts a force equal in magnitude (but) opposite in direction (on A).

Applications of Newton’s laws of motion: Seat belts / Rocket travel / Ball games

To Show that F = ma is a special case of Newton’s Second Law

From Newton II: Force is proportional to the rate of change of momentum

Force ( rate of change of momentum

F ( (mv – mu)/t

F ( m(v-u)/t

F ( ma

F = k (ma)

F = ma

Note: k = 1 because of how we define the newton (a force of 1 N gives a mass of 1 kg an acceleration of 1 m s-2)*

The Principle of Conservation of Momentum

states that in any collision between two objects, the total momentum before impact equals total momentum after impact, provided no external forces act on the system.

(If you forget the bit in italics you lose half marks!)

In symbols

Note that if the two objects coalesce (stick together) after collision,

there is only one final velocity, and the above equation becomes

Areas where the principle of conservation of momentum applies:

• Collisions (ball games)

• Acceleration of aircraft

• Jet aircraft

Mandatory Experiments

• To show that the acceleration of a body is proportional to the force acting on it.

• To Verify the Principle of Conservation of Momentum.

Leaving Cert Physics Syllabus

|Content |Depth of Treatment |Activities |STS |

| | | | |

|1.Newton’s laws of motion |Statement of the three laws. |Demonstration of the three laws using |Applications: |

| | |air track or tickertape timer etc. |seat belts |

| | | |rocket travel. |

| | | |Sports, all ball games. |

| | | | |

| |Force and momentum: definitions and units. | |Important of friction in everyday |

| |Vector nature of forces to be stressed. | |experience, e.g. walking, use of |

| |F = ma as a special case of Newton’s second |Appropriate calculations. |lubricants etc. |

| |law. | | |

| |Friction: a force opposing motion. | | |

| | | | |

|2. Conservation of momentum |Principle of conservation of momentum. |Demonstration by any one suitable |Collisions (ball games), acceleration |

| | |method. |of spacecraft, jet aircraft. |

| | |Appropriate calculations. | |

TO SHOW THAT ACCELERATION IS PROPORTIONAL TO THE FORCE WHICH CAUSED IT

APPARATUS

Set of weights, electronic balance, trolley, ticker-tape timer and tape.

DIAGRAM

[pic]

PROCEDURE

1. Set up the apparatus as shown in the diagram.

2. Start by taking one weight from the trolley and adding it to the hanger at the other end.

3. Note the weight at this end (including the weight of the hanger) using an electronic balance.

4. Release the system which allows the trolley to accelerate down the track.

5. Use the ticker-tape timer to calculate the acceleration.

6. Repeat these steps about seven times, each time taking a weight from the trolley and adding it to the other end.

7. Record the results for force and acceleration in a table.

8. Draw a graph of Force (on the y-axis) against acceleration (on the x-axis). The slope of the graph corresponds to the mass of the system (trolley plus hanger plus all the weights)

RESULLTS

|Force |Acceleration (m s-2)|

|(N) | |

| | |

| | |

| | |

| | |

CONCLUSION

Our graph resulted in a straight line through the origin, verifying that the acceleration is proportional to the force, as the theory predicted.

The slope of our force-acceleration graph was 0.32, which was in rough agreement with the mass of the system which we measured to be 0.35 kg.

PRECAUTIONS / SOURCES OF ERROR

1. When adding weights to the hanging masses, you must take them from on top of the trolley.

2. Ensure that the runway in smooth, free of dust, and does not sag in the middle.

3. Ensure that the runway is tilted just enough for the trolley to roll at constant speed when no force is applied.

When must the hanging weights be taken from on top of the trolley?

Answer: so that the mass of the system can be kept constant

We’re looking to investigate the relationship between the acceleration of an object and the force which caused it.

The force which is causing the acceleration is the hanging weights. What mass is accelerating as a result of these weights dropping?

Well obviously the trolley plus the weights sitting on it are accelerating, but not just that; the hanging weights themselves are also accelerating, so the total mass accelerating as a result of the hanging weights is:

trolley + weights sitting on trolley + hanging weights

Now if we’re looking to investigate the relationship between the acceleration of an object and the force which caused it we need to keep all other variables constant. In this case one other variable is the mass which is )being accelerated. The only way to increase the hanging weights while keeping the mass of the system constant, is to transfer weights from the trolley to the hanging weights.

Using the ticker-tape system

If using the ticker-tape you will need to calculate the velocity at the beginning (u = s1/ t1), the velocity at the end (v = s2/ t2) and then use the equation v2 = u2 + 2as, where s is the distance between the middle of first set of dots and the middle of the second set of dots.

|s1 (m) |t1(s) |u (m s -1) |s2 (m) |

| | | | |

| | | | |Total Before| | | |

|acceleration /m s–2 |0.10 |0.22 |0.32 |0.44 |0.55 |0.65 |0.76 |

i) Draw a labelled diagram of the apparatus used in the experiment.

ii) Outline how the student measured the applied force.

iii) Plot a graph, on graph paper of the acceleration against the applied force. Put acceleration on the horizontal axis (X-axis).

iv) Calculate the slope of your graph and hence determine the mass of the body.

v) Give one precaution that the student took during the experiment.

1. [2006 OL]

In a report of an experiment to verify the principle of conservation of momentum, a student wrote the following:

I assembled the apparatus needed for the experiment. During the experiment I recorded the mass of the trolleys and I took measurements to calculate their velocities. I then used this data to verify the principle of conservation of momentum.

i) Draw a labelled diagram of the apparatus used in the experiment.

ii) How did the student measure the mass of the trolleys?

iii) Explain how the student calculated the velocity of the trolleys.

iv) How did the student determine the momentum of the trolleys?

v) How did the student verify the principle of conservation of momentum?

2. [2005]

In an experiment to verify the principle of conservation of momentum, a body A was set in motion with a constant velocity. It was then allowed to collide with a second body B, which was initially at rest and the bodies moved off together at constant velocity.

The following data was recorded.

Mass of body A = 520.1 g

Mass of body B = 490.0 g

Distance travelled by A for 0.2 s before the collision = 10.1 cm

Distance travelled by A and B together for 0.2 s after the collision = 5.1 cm

i) Draw a diagram of the apparatus used in the experiment.

ii) Describe how the time interval of 0.2 s was measured.

iii) Using the data calculate the velocity of the body A before and after the collision.

iv) Show how the experiment verifies the principle of conservation of momentum.

v) How were the effects of friction and gravity minimised in the experiment?

Exam solutions

1. A Force is anything which can cause an object to accelerate.

2. A force of 1 N gives a mass of 1 kg an acceleration of 1 m s-2.

3. Newton’s first law of motion states that an object stays at rest or moves with constant velocity unless an external force acts on it.

4. There are no external forces acting on the spacecraft so from Newton’s 1st law of motion the object will maintain its velocity.

5. Newton’s Second Law of Motion states that the rate of change of an object’s momentum is directly proportional to the force which caused it, and takes place in the direction of the force.

6. Newton’s First Law of Motion states that every object will remain in a state of rest or travelling with a constant velocity unless an external force acts on it.

7. Newton’s laws of motion: see above

8. F = ma ( F = 150 × 7 = 1050 N

9. Gravity (or weight), friction, air resistance.

10. W = mg = 120 × 1.6 = 192 N.

11. Because acceleration due to gravity is greater on the earth (because the mass of the earth is greater than the mass of the moon).

12. Because gravity is less on the moon.

13.

i) W = mg = 2000 × 9.8 = 19600 N

ii) 2000 kg

iii) W = mg = 2000 × 1.6 = 3200 N

iv) The force of gravity is less on moon so less force is needed to escape.

14.

i) It accelerates upwards.

ii) It accelerates forward.

iii) v = u + at ⇒ 0 = 60 + a (120) ⇒ a = - 0.5 m s-2

iv) F = ma ⇒ F = 50 000 × 0.5 = 25 000 N.

v) They would continue to move at the greater initial velocity and so would be ‘thrown’ forward.

15. Weight (W) downwards; reaction (R) upwards; force to left (due to friction or curled fingers)

16. F = ma = 90 × 0.83 = 75 N Down

17. Weight acting down.

Air resistance / friction / buoyancy acting up.

Air resistance = weight, therefore resultant force = 0

Therefore acceleration = 0

18. v2 = u2 + 2as ( 0 = (2.48)2 + 2a(2)

a = 1.56 m s-2

F = ma = (0.008)(1.6) = 0. 0.013 N

19.

i) v2= u2 + 2as ( (12.2)2 = 0 +2a(25) ( a = 2.98 m s–2

ii) W = mgsin( = mgsin20 = 234.63 N

iii) Force down (due to gravity) – Resistive force (due to friction) = Net force

Force down (due to gravity) = 234.63 N

Net force= 70(2.98) = 208.38 N

Friction force = 234.63 – 208.38 = 26.25 N

iv) v2= u2 + 2as ( u2 = 2g(s) ( s = 7.59 m

v) Graph: velocity on vertical axis, time on horizontal axis, with appropriate numbers on both axes.

20. If the wheelchair is moving at constant speed then the force up must equal the force down, so to calculate the size of the force up, we just need to calculate the force down:

F = mg Sin(

= 900 Sin 10o

= 156.3 N

21. Friction is a force which opposes the relative motion between two objects.

22.

i) v = u + at

50 = 0 + 0.5t

t = 50/0.5 = 100 s

ii) s = ut + ½ at2 (but a = 0)

s = 50 × (90×60) = 270000 m

iii) v2 = u2 + 2as

0 = 502 + 2a(500)

a = −2500/1000 = − 2.5 m s-1

iv) F = ma

F = 30000× (−)(2.5) = - 75000 N = 75 kN

v) A = friction/retardation / resistance to motion

B = weight / force of gravity

vi) The train will move at constant speed.

vii) See diagram

23.

i) Fnet = ma = (750)(1.2) = 900 N east.

ii) Fnet = Fcar - Ffriction

900 = 2000 - Ffriction ⇒ Ffriction = 1100 N west

iii) Friction causes deceleration: a = F ÷ m

a = (-1100) ÷ 750 = - 1.47 ms-2

v 2 = u 2 + 2as

0 = 25 +2(-1.47) s or s = 213 m

Momentum

24. Momentum is the defined as the product of mass multiplied by velocity.

25. The unit of momentum is the kg m s-1

25. The principle of conservation of momentum states that in any collision between two objects, the total momentum before impact equals total momentum after impact, provided no external forces act on the system.

26. From Newton II: Force is proportional to the rate of change of momentum

F ( (mv – mu)/t ⇒F ( m(v-u)/t ⇒F ( ma ⇒F = k (ma) ⇒F = ma

27. The gas moves down (with a momentum) causing the rocket to move up (in the opposite direction with an equal momentum)

28. Momentum = mass × velocity = 5 × 10 = 50 kg m s-1.

29.

i) (mu = ) 12 × 3.5 = 42 kg m s-1

ii) Momentum before = Momentum after

42 = m3v3 ( v3 = 42/m3 ( v = 42/24 = 1.75 ( m s-1)

30. m1u1 + m2u2 = m1v1 + m2v2 ( 0 = (40)(2) + (50)x

( x = - 1.6 m s-1.

31.

i) m1u1 + m2u2 = m1v1 + m2v2

(50000 × 2) = (50000 × 0.5) + (50m)

m =1500 kg

ii) In what direction should the gas be expelled?

Forward (toward the space station).

iii) Explain how the principle of conservation of momentum is applied to changing the direction in which a spacecraft is travelling.

As the gas is expelled in one direction the rocket moves in the other direction.

32. m1u1 + m2u2 = m1v1 + m2v2

(0.01)(2) = (0.008) v2

v2= 2.48 m s-1

33. From Newton II: Force ( rate of change of momentum

F ( (mv – mu)/t

F = (mv – mu)/t {proportional constant = 1}

(mv – mu) = F × t = (9 × 103)( 0.6 × 10-3) = 5.4 kg m s-1.

Mandatory experiments

34.

i) See diagram in next question.

ii) Tilt the runway slightly, oil the track.

iii) By weighing the masses and hanger on an electronic balance.

iv) See graph

v) Acceleration is directly proportional to the applied force.

35.

i) See diagram.

ii) Outline how the student measured the applied force.

The applied force corresponds to the weight of the hanger plus weights; the value of the weights is written on the weights themselves.

iii) Plot a graph, on graph paper of the acceleration against the applied force. Put acceleration on the horizontal axis (X-axis).

See graph.

iv)

Substituting in two values (from the graph, not the table) should give a slope of approximately 0.9.

This means that the mass = 0.9 kg.

v) Oil the trolley wheels, dust the runway, oil the pulley.

36.

i) See diagram

ii) By using an electronic balance.

iii) By taking a section of the tape and using the formula velocity = distance/time. We measured the distance between 11 dots and the time was the time for 10 intervals, where each interval was 1 50th of a second.

iv) Using the formula momentum = mass × velocity.

v) By calculating the total momentum before and afterwards and showing that the total momentum before = total momentum after.

37.

i) See diagram

ii) It corresponded to 10 intervals on the ticker-tape.

iii) Velocity before: v = s/t = 0.101/0.2

v = 0.505 m s-1 ≈ 0.51 m s-1

Velocity after: v = 0.051/0.2

v = 0.255 m s-1 ≈ 0.26 m s-1

iv) Momentum before:

p = mv = (0.5201)(0.505) = 0.263 ≈ 0.26 kg m s-1

Momentum after:

p = mv = (0.5201 + 0.4900)(0.255)

p = 0.258 ≈ 0.26 kg m s-1

Momentum before ≈ momentum after

v) Friction: sloped runway // oil wheels or clean track

Gravity: horizontal track // frictional force equal and // tilt track so that trolley moves with constant velocity

[2010 OL]

You carried out an experiment to investigate the relationship between the acceleration of a body and the force applied to it.

You did this by applying a force to a body and measuring the resulting acceleration.

The table shows the data recorded during the experiment.

Force / N |0.20 |0.25 |0.30 |0.35 |0.40 |0.45 |0.50 | |acceleration / m s−2 |0.4 |0.5 |0.6 |0.7 |0.8 |0.9 |1.0 | |

i) Draw a labelled diagram of the apparatus you used

Labelled diagram to show:

Trolley / rider

Runway / air-track

Means of applying a force e.g. string over pulley to weight on pan

Means of measuring acceleration e.g. 2 photo-gates (and timer) // tickertape (and timer)

ii) How did you measure the applied force?

Weighed the mass (and pan) / mg // from the (digital Newton) balance

iii) How did you minimise the effect of friction during the experiment?

Slant/clean the runway // oil (the trolley) wheels / frictionless wheels

iv) Plot a graph on graph paper of the body’s acceleration against the force applied to it

v) What does your graph tell you about the relationship between the acceleration of the body and the force applied to it?

They are proportional.

[2010 OL]

i) An ice skater of mass 50 kg was moving with a speed of 6 m s−1 then she collides with another skater of mass 70 kg who was standing still. The two skaters then moved off together.

Calculate the momentum of each skater before the collision?

50 × 6 = 300 kg m s−1

70 × 0 = 0 kg m s−1

ii) What is the momentum of the combined skaters after the collision?

300 kg m s−1

iii) Calculate the speed of the two skaters after the collision.

300 = (50 + 70) v

v = 2.5 m s−1

[2010 OL]

A cyclist on a bike has a combined mass of 120 kg.

The cyclist starts from rest and by pedalling applies a net force of 60 N to move the bike along a horizontal road.

Calculate the acceleration of the cyclist

F = ma, a = 60/120

a = 0.5 m s–2

[2010 OL]

A cyclist stops peddling and continues to freewheel before coming to a stop. Why does the bike stop?

Due to friction / air resistance.

[2010]

In an experiment to investigate the relationship between the acceleration of a body and the force applied to it, a student recorded the following data.

F/N |0.20 |0.40 |0.60 |0.80 |1.00 |1.20 |1.40 | |a/m s–2 |0.08 |0.18 |0.28 |0.31 |0.45 |0.51 |0.60 | |

i) Describe the steps involved in measuring the acceleration of the body.

Measure/calculate the initial velocity/speed

Measure/calculate the velocity/speed again (t seconds later)

Measure time interval from initial to final velocities / distance between light gates

Use relevant formula

Datalogging method:

Align motion sensor with body (e.g. trolley) / diagram

Select START and release body

(Select STOP and) display GRAPH of ‘a vs. t’ // ‘v vs. t’

(Use tool bar to) find average value for a // use slope (tool) to find a (= dv /dt)

ii) Using the recorded data, plot a graph to show the relationship between the acceleration of the body and the force applied to it.

Label axes correctly on graph paper

Plot six points correctly

Straight line

Good distribution

iii) What does your graph tell you about this relationship?

Acceleration is proportional to the applied force.

iv) Using your graph, find the mass of the body.

The mass of the body corresponds to the slope of the graph = 2.32 kg [range: 2.1 - 2.4 kg]

v) On a trial run of this experiment, a student found that the graph did not go through the origin.

Suggest a reason for this and

Friction / dust on the track slowing down the trolley.

vi) Describe how the apparatus should be adjusted, so that the graph would go through the origin.

Elevate/adjust the track/slope

-----------------------

F = ma

W = mg

( = mv

m1u1 + m2u2 = m1v1 + m2v2

m1 u1 + m2 u2 = (m1 + m2)v3

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